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Electromagnetic and Electromechanical
Engineering Principles
Notes 02
Magnetics and Energy Conversion
Marc T. Thompson, Ph.D.
Thompson Consulting, Inc.
9 Jacob Gates Road
Harvard, MA 01451
Phone: (978) 456-7722
Email: [email protected]
Website: http://www.thompsonrd.com
© Marc Thompson, 2006-2008
Portions of these notes reprinted from Fitzgerald, Electric Motors, 6th edition
Electromechanics
Magnetics and Energy Conversion
2-1
Course Overview --- Day 2
Electromechanics
Magnetics and Energy Conversion
2-2
Overview of Magnetics
• Review of Maxwell’s equations
• Ampere’s law, Gauss’ law, Faraday’s law
• Magnetic circuits
• Flux, flux linkage, inductance and energy
Electromechanics
Magnetics and Energy Conversion
2-3
Review of Maxwell’s Equations
• First published by James Clerk Maxwell in
1864
• Maxwell’s equations couple electric fields
to magnetic fields, and describe:
– Magnetic fields
– Electric fields
– Wave propagation (through the wave
equation)
• There are 4 Maxwell’s equations, but in
magnetics we generally only need 3:
– Ampere’s Law
– Faraday’s Law
– Gauss’ Magnetic Law
Electromechanics
James Clerk Maxwell
Magnetics and Energy Conversion
2-4
Review of Maxwell’s Equations
• We’ll review Maxwell’s equations in words, followed by a
little bit of mathematics and some computer simulations
showing the magnetic fields
Electromechanics
Magnetics and Energy Conversion
2-5
Ampere’s Law
• Flowing current creates a magnetic
field
r v
r v
v v d
∫C H ⋅ dl = ∫S J ⋅ dA + dt ∫S ε o E ⋅ dA
• In magnetic systems, generally there
is high current and low voltage (and
hence low electric field) and we can
approximate for low d/dt:
r v
v v
∫ H ⋅ dl ≈ ∫ J ⋅ dA
C
S
• In words: the magnetic flux density
integrated around any closed contour
equals the net current flowing through
the surface bounded by the contour
Electromechanics
André-Marie Ampère
Magnetics and Energy Conversion
2-6
Finite-Element Analysis (FEA)
• Very useful tool for visualizing and solving shapes and
magnitudes of magnetic fields
• FEA is often used to simulate and predict the performance
of motors, etc.
• Following we’ll see some 2-dimensional (2D) FEA results to
help explain Maxwell’s equations
Electromechanics
Magnetics and Energy Conversion
2-7
Field From Current Loop, NI = 500 A-turns
• Coil radius R = 1”; plot from 2D finite-element analysis
Electromechanics
Magnetics and Energy Conversion
2-8
Faraday’s Law
• A changing magnetic flux impinging on
a conductor creates an electric field
and hence a current (eddy current)
r v
d r v
∫C E ⋅ dl = − dt ∫S B ⋅ dA
• The electric field integrated around a
closed contour equals the net timevarying magnetic flux density flowing
through the surface bound by the
contour
• In a conductor, this electric field creates
r
r
a current by:
J = σE
• Induction motors, brakes, etc.
Electromechanics
Michael Faraday
Magnetics and Energy Conversion
2-9
Circular Coil Above Conducting Aluminum Plate
• Flux density plots at DC and 60 Hz
• At 60 Hz, currents induced in plate via magnetic induction
create lift force
DC
Electromechanics
60 Hz
Magnetics and Energy Conversion
2-10
Demonstration of Faraday’s Law: Electrodynamic
Drag (NdFeB Magnet-in-Tube)
• Process:
– Moving magnet creates changing magnetic field in
copper tube
– Changing magnetic field creates induced voltage
– Induced voltage creates current
– By Lorentz force law, induced current and applied
magnetic field create drag force
Electromechanics
Magnetics and Energy Conversion
2-11
Gauss’ Magnetic Law
• Gauss' magnetic law says that the
integral of the magnetic flux density over
any closed surface is zero, or:
∫ B ⋅ dA = 0
S
• This law implies that magnetic fields are
due to electric currents and that
magnetic charges (“monopoles”) do not
exist.
• Note: similar form to KCL in circuits.
(We’ll use this analogy later…)
B1, A1
B2, A2
B3, A3
B3 A3 = B1 A1 + B2 A2
Carl Friedrich Gauss
Electromechanics
Magnetics and Energy Conversion
2-12
Gauss’ Law --- Continuity of Flux Lines
φ1 + φ2 + φ3 = 0
Reference: N. Mohan, et. al., Power Electronics Converters, Applications and Design, Wiley, 2003, pp. 48
Electromechanics
Magnetics and Energy Conversion
2-13
Lorentz Force Law
• Experimentally derived rule:
F = ∫ J × BdV
• For a wire of length l carrying current I perpendicular to
a magnetic flux density B, this reduces to:
F = IBl
Electromechanics
Magnetics and Energy Conversion
2-14
Lorentz Force Law and the Right Hand Rule
F = ∫ J × BdV
Reference: http://www.physics.brocku.ca/faculty/sternin/120/slides/rh-rule.html
Electromechanics
Magnetics and Energy Conversion
2-15
Intuitive Thinking about Magnetics
• By Ampere’s Law, the current J and the magnetic field H
are generally at right angles to one another
• By Gauss’ law, magnetic field lines loop around on
themselves
– No magnetic monopole
• You can think of high- μ magnetic materials such as steel
as an easy conduit for magnetic flux…. i.e. the flux easily
flows thru the high- μ material
Electromechanics
Magnetics and Energy Conversion
2-16
Magnetic Field (H) and Magnetic Flux Density (B)
• H is the magnetic field (A/m in SI units) and B is the
magnetic flux density (Weber/m2, or Tesla, in SI units)
• B and H are related by the magnetic permeability μ by B =
μH
• Magnetic permeability μ has units of Henry/meter
• You can think of high- μ magnetic materials such as steel
as an easy conduit for magnetic flux…. i.e. the flux easily
flows thru the high- μ material
• In free space μo = 4π×10-7 H/m
• Note that B and H are vectors; they have both a magnitude
and a direction
Electromechanics
Magnetics and Energy Conversion
2-17
Right Hand Rule and Direction of Magnetic Field
Reference: http://sol.sci.uop.edu/~jfalward/magneticforcesfields/magneticforcesfields.html
Electromechanics
Magnetics and Energy Conversion
2-18
Forces Between Current Loops
Reference: http://sol.sci.uop.edu/~jfalward/magneticforcesfields/magneticforcesfields.html
Electromechanics
Magnetics and Energy Conversion
2-19
Inductor Without Airgap
• Magnetic flux is constrained to flow within steel
Constitutive relationships
In free space:
B = μo H
Magnetic permeability of free space μo = 4π×10-7 Henry/meter.
In magnetic material, magnetic permeability is higher than μo:
B = μH
Electromechanics
Magnetics and Energy Conversion
2-20
Example: Flux in Inductor with Airgap
Ampere's law:
∫ H ⋅ dl = ∫ J ⋅ dA ⇒ H l
c c
+ H g g = NI
S
Let's use constitutive relationships:
Bc
μc
Electromechanics
lc +
Bg
μo
g = NI
Magnetics and Energy Conversion
2-21
Example: Flux in Inductor
Φ with Airgap
Bc =
Ac
Φ
Ac
Put this into previous expression:
⎛ lc
g ⎞⎟
⎜
NI = Φ
+
⎜ μA μ A ⎟
o g ⎠
⎝ c
Solve for flux:
NI
Φ=
⎛ lc
g ⎞⎟
⎜
+
⎜ μA μ A ⎟
o g ⎠
⎝ c
If g/μoAg >> lc/μAc,
NI
Φ≈
g
μ o Ag
Bg =
Electromechanics
Magnetics and Energy Conversion
2-22
Magnetic Circuits
• Use Ohm’s law analogy to model magnetic circuits
V ⇔ NI
I ⇔Φ
R⇔ℜ
• Use magnetic “reluctance” instead of resistance
l
l
⇔ℜ =
R=
σA
μA
• This is a very powerful method to get approximate
answers in magnetic circuits
Electromechanics
Magnetics and Energy Conversion
2-23
Magnetic-Electric Circuit Analogy
• In an electric circuit, voltage V forces current I to flow
through resistances R
• In a magnetic circuit, MMF NI forces flux Φ to flow through
reluctances ℜ
Electromechanics
Magnetics and Energy Conversion
2-24
C-Core with Gap --- Using Magnetic Circuits
• Flux in the core is easily found by:
Φ=
NI
NI
=
lp
g
ℜ core + ℜ gap
+
μ c Ac μ o Ac
N turns
Airgap, g
• Now, note what happens if g/μo >>
lp/μc: The flux in the core is now
approximately independent of the
core permeability, as:
NI
NI
Φ≈
≈
g
ℜ gap
μ o Ac
NΦ
N
N
≈
≈
• Inductance: L =
g
ℜ gap
I
μ o Ac
2
Electromechanics
Average magnetic path length lp
inside core,
cross-sectional area A
2
Φ
ℜ core
NI
+
-
ℜ gap
Magnetics and Energy Conversion
2-25
C-Core with Gap --- FEA
Electromechanics
Magnetics and Energy Conversion
2-26
Fringing Fields in Airgap
• If fringing is negligible, Ac = Ag
Electromechanics
Magnetics and Energy Conversion
2-27
Example: C-Core with Airgap
• Fitzgerald, Example 1.1; with Bc = 1.0T, find reluctances,
flux and coil current
Electromechanics
Magnetics and Energy Conversion
2-28
Example: C-Core with Airgap
Electromechanics
Magnetics and Energy Conversion
2-29
Example: C-Core with Airgap --- FEA
• NI = 400 A-turns
Electromechanics
Magnetics and Energy Conversion
2-30
Example: C-Core with Airgap --- FEA
• NI = 400 A-turns, close up near the core
Electromechanics
Magnetics and Energy Conversion
2-31
Example: C-Core with Airgap --- FEA Result
• Flux density in the core is approximately 1 Tesla
Electromechanics
Magnetics and Energy Conversion
2-32
Example: C-Core with Airgap --- Gap Detail
Electromechanics
Magnetics and Energy Conversion
2-33
Example: Simple Synchronous Machine
• Fitzgerald, Example 1.2
• Assuming μÆ∞, find airgap flux Φ and flux density Bg
Assume I = 10A, N = 1000 turns, g = 1 cm and Ag = 2000
cm2
Electromechanics
Magnetics and Energy Conversion
2-34
Aside: What Does Infinite Core μ Imply?
In core:
Φ c = Bc AC
In airgap:
Φ g = B g Ag
In core, Bc is finite; this means that if μÆ ∞, then Hc Æ 0 for finite
Bc. Also, infinite μ implies zero reluctance
Electromechanics
Magnetics and Energy Conversion
2-35
Example: Simple Synchronous Machine
• Some initial thoughts,
before doing any
equations:
– By symmetry, airgap
flux and flux density
are the same in both
gaps
– Since permeability is
infinite, H inside steel
is zero
Electromechanics
Magnetics and Energy Conversion
2-36
Example: Simple Synchronous Machine
Reluctances:
ℜ g1 = ℜ g 2 =
g
A − turns
(0.01)
=
=
39789
μ o Ag (4π × 10 −7 )(2000)(0.012 )
Wb
Flux:
Φ=
NI
(1000)(10)
=
= 0.126 Wb
ℜ g1 + ℜ g 2 (2)(39789)
Magnetic flux density:
0.126 Wb
Φ
Wb
B=
=
=
0
.
63
= 0.63 T
2
2
Ag (2000)(0.01 )
m
Electromechanics
Magnetics and Energy Conversion
2-37
Flux Linkage, Voltage and Inductance
• By Faraday’s law, changing magnetic flux density creates
an electric field (and a voltage)
r v
d r v
∫C E ⋅ dl = − dt ∫S B ⋅ dA
• Induced voltage:
dΦ dλ
v=N
=
dt
dt
λ = "flux linkage" = NΦ
Inductance relates flux linkage to current
L=
λ
I
Electromechanics
Magnetics and Energy Conversion
2-38
Finding Inductance Using Magnetic Circuits
• Let’s at first assume infinite core permeability; this means
that the reluctance of the core is zero
• Note that inductance always scales as N2 (why is that?)
NI
Φ≈
g
μ o Ag
λ ≈ NΦ =
L≈
Electromechanics
λ
I
=
μ o Ag N 2 I
g
μ o Ag N 2
g
Magnetics and Energy Conversion
2-39
Magnetic Circuit with Two Airgaps
Total flux: Φ =
NI
ℜ1 ℜ 2
Reluctances: ℜ 1 =
g1
g2
and ℜ 2 =
μ o A1
μ o A2
Inductance:
⎛ ℜ + ℜ2
λ NΦ
= N 2 ⎜⎜ 1
L= =
I
I
⎝ ℜ 1ℜ 2
Electromechanics
⎛A
⎞
A ⎞
⎟⎟ = μ o N 2 ⎜⎜ 1 + 2 ⎟⎟
⎠
⎝ g1 g 2 ⎠
Magnetics and Energy Conversion
2-40
Magnetic Circuit with Two Airgaps (cont.)
Flux in leg#1: Φ 1 =
NI μ o A1 NI
=
ℜ1
g1
Flux in leg#2: Φ 2 =
NI μ o A2 NI
=
ℜ2
g2
Φ 1 μ o NI
=
A1
g1
Φ 2 μ o NI
=
Flux density in leg#2: B2 =
A2
g2
Flux density in leg#1: B1 =
Electromechanics
Magnetics and Energy Conversion
2-41
Inductance vs. Relative Permeability
• What happens if we assume finite core permeability?
• Reluctance of the core is now finite as well
lc
g
and ℜ g =
μAc
μo A
Let’s assume that Ac = Ag = A
Reluctances: ℜ c =
Flux: Φ =
NI
ℜc + ℜ g
N 2I
Flux linkage: λ = NΦ =
ℜc + ℜ g
N 2 μo A
N2
N2
N= =
=
=
Inductance:
I ℜ c + ℜ g ⎛ lc ⎞ ⎛ g ⎞
⎛l μ
⎟⎟ g + ⎜⎜ c o
⎜⎜
⎟⎟ + ⎜⎜
⎝ μ
⎝ μA ⎠ ⎝ μ o A ⎠
Inductance is independent of core permeability if:
l μ
g >> c o
λ
μ
Electromechanics
Magnetics and Energy Conversion
2-42
⎞
⎟⎟
⎠
Inductance vs. Relative Permeability
Electromechanics
Magnetics and Energy Conversion
2-43
Example: Effects of Finite Permeability
• Fitzgerald, problem 1.5
Electromechanics
Magnetics and Energy Conversion
2-44
Example: Effects of Finite Permeability
• Relative μ vs. B
Electromechanics
Magnetics and Energy Conversion
2-45
Example: Effects of Finite Permeability
• B/H curve
Electromechanics
Magnetics and Energy Conversion
2-46
Example: Effects of Finite Permeability
• Current calculation
From Ampere’s law:
H c l c + H g g = NI
In core:
B
Hc = c
μ
In gap (let’s assume Bc = Bg = B):
B
Hg =
μo
Put this back into Ampere’s law:
Bl c Bg
+
= NI
μ
μo
⎛ B ⎞⎛ l c
⎞
⎟⎟⎜⎜
∴ I = ⎜⎜
+ g ⎟⎟ = 65.8 A
⎠
⎝ μ o N ⎠⎝ μ r
Electromechanics
Magnetics and Energy Conversion
2-47
Example: Effects of Finite Permeability
• Coil flux linkage λ as a function of coil current
• Note that at low current, λ-I curve is linear, indicating
constant inductance
Electromechanics
Magnetics and Energy Conversion
2-48
Example:
MATLAB Script
Electromechanics
Magnetics and Energy Conversion
2-49
Inductance and Energy
• Magnetic stored energy (in Joules) is:
1 2
W = LI
2
• This is a good thing to remember
Electromechanics
Magnetics and Energy Conversion
2-50
Magnetic Circuit with Two Windings
• Note that flux is the sum of flux due to i1 and that due to i2
Electromechanics
Magnetics and Energy Conversion
2-51
Magnetic Circuit with Two Windings
• Note that by the right-hand rule the flux due to i1 and i2 are
additive given the current directions shown
Electromechanics
Magnetics and Energy Conversion
2-52
Magnetic Circuit with Two Windings
• Note that by the right-hand rule the flux due to i1 and i2
are additive given the current directions shown
⎛ μ o Ac
Flux: Φ = ( N 1i1 + N 2 i 2 )⎜⎜
⎝ g
⎞
⎟⎟
⎠
Flux linkage for coil #1:
⎛μ A ⎞
⎛μ A
λ1 = N 1 Φ = N 12 i1 ⎜⎜ o c ⎟⎟ + N 1 N 2 i 2 ⎜⎜ o c
⎝ g ⎠
⎝ g
⎞
⎟⎟
⎠
We can write this as:
λ1 = L11 i1 + L12 i 2
L11 is “self inductance” of coil #1
L12 is “mutual inductance” between coils #1 and
#2
Electromechanics
Magnetics and Energy Conversion
2-53
Magnetic Circuit with Two Windings
Flux linkage for coil #2:
⎛ μ o Ac
λ 2 = N 2 Φ = N 1 N 2 i1 ⎜⎜
⎝ g
⎞
2 ⎛ μ o Ac
⎟⎟ + N 2 i2 ⎜⎜
⎠
⎝ g
Or rewriting:
λ2 = L21i1 + L22 i2
Or, in matrix form:
⎡ λ1 ⎤ ⎡ L11 L12 ⎤ ⎡ i1 ⎤
⎢λ ⎥ = ⎢ L
⎥ ⎢i ⎥
L
22 ⎦ ⎣ 2 ⎦
⎣ 2 ⎦ ⎣ 21
Electromechanics
Magnetics and Energy Conversion
2-54
⎞
⎟⎟
⎠
“Soft” Magnetic Materials
• Materials with a small B/H curve, such as steels, etc.
• Much of the previous analysis assumed that steel had
infinite permeability (μ Æ ∞) or that permeability was
constant and large.
• However, soft magnetic materials exhibit both saturation
and losses
Electromechanics
Magnetics and Energy Conversion
2-55
B-H Curve and Saturation
• Definition of magnetic permeability: slope of B-H curve
Reference: N. Mohan, et. al., Power Electronics Converters, Applications and Design, Wiley, 2003, pp. 48
Electromechanics
Magnetics and Energy Conversion
2-56
BH Curve for M-5 Steel
• Note horizontal scale is logarithmic
Electromechanics
Magnetics and Energy Conversion
2-57
Hysteresis Loop
• Real-world magnetic materials have a “hysteresis loop”
• Hysteresis loss is proportional to shaded area
Electromechanics
Magnetics and Energy Conversion
2-58
B-H Loop for M-5 Grain-Oriented Steel
• Only the top half of the loops shown for steel 0.012” thick
Electromechanics
Magnetics and Energy Conversion
2-59
BH Curves for Various Soft Magnetic Materials
Reference: E. Furlani, Permanent Magnet and Electromechanical Devices, Academic Press, 2001, pp. 41
Electromechanics
Magnetics and Energy Conversion
2-60
Relationship Between Voltage, Flux and Current
Electromechanics
Magnetics and Energy Conversion
2-61
Exciting RMS VA per kg at 60 Hz
Electromechanics
Magnetics and Energy Conversion
2-62
Hysteresis Loop Size Increases with Frequency
• Hysteresis loss increases as frequency increases
Reference: Siemens, Soft Magnetic Materials (Vacuumschmelze Handbook), pp. 30
Electromechanics
Magnetics and Energy Conversion
2-63
Total Core Loss
• Total core loss is the sum of:
– Hysteresis loss
– Eddy current losses
• Eddy current losses are due to induced currents (via
Faraday’s law)
• Eddy current losses are minimized by laminating magnetic
cores
Electromechanics
Magnetics and Energy Conversion
2-64
Laminated Core
Cores made from conductive magnetic materials must be
made of many thin laminations. Lamination thickness < skin
depth.
•
0.5 t
t (typically
0.3 mm)
Insulator
Magnetic steel
lamination
Electromechanics
Magnetics and Energy Conversion
2-65
Eddy Current Loss in Lamination
w
x
z
d
y
L
dx
B sin( ωt)
x
-x
Eddy current flow path
Electromechanics
Magnetics and Energy Conversion
2-66
Total Core Loss
• M-5 steel at 60 Hz
Electromechanics
Magnetics and Energy Conversion
2-67
Total Core Loss vs. Frequency and Bmax
• Core loss depends on peak flux density and excitation
frequency
• This is the curve for a high frequency core material
Reference: http://www.jfe-steel.co.jp/en/products/electrical/jnhf/02.html
Electromechanics
Magnetics and Energy Conversion
2-68
Process to Find Core Loss
• Find maximum B
• From this B and
switching frequency,
find core loss per kg
• Total loss is power
density × mass of
core
Electromechanics
Magnetics and Energy Conversion
2-69
Example: C-Core with Airgap --- Current
• Fitzgerald, Example 1.7; find the current necessary to
produce Bc = 1T
Electromechanics
Magnetics and Energy Conversion
2-70
Example: C-Core with Airgap --- Solution
The value of Hc needed for Bc = 1 Tesla is read from the
chart:
H c = 11 A − turns / meter
The MMF drop in the core is:
H c l c = (11)(0.3) = 33 A − turns
The MMF drop in the airgap is:
B g g (1.0)(5 × 10 −4 )
Hgg =
=
= 396 A − turns
−7
μo
4π × 10
The winding current is:
MMF 33 + 396
∑
I=
=
= 0.8 A
N
500
Electromechanics
Magnetics and Energy Conversion
2-71
Example: Inductor
• Fitzgerald, Example 1.8
Material: M-5 steel
f = 60 Hz
N=200
Bc = 1.5 sinωt Tesla
Steel is 94% of cross
section
Density of steel = 7.65
g/cm3
Find:
(a) Applied voltage
(b) Peak current
(c) RMS current
(d) Core loss
Electromechanics
Magnetics and Energy Conversion
2-72
Example: Excitation Voltage
From Faraday’s law:
dBc
dΦ
dλ
=N
= NAc
e=
dt
dt
dt
Ac = 2in × 2in × 0.94 = 3.76in = 2.4 × 10
2
−3
m
2
dBc
= (1.5ω ) cos(ωt ) = 565 cos(ωt )
dt
e = (200)(2.4 × 10 −3 )(565 cos(ωt )) = 274 cos(ωt )
Electromechanics
Magnetics and Energy Conversion
2-73
Example: Peak Current in Winding
From Figure 1.10, B = 1.5T requires H =
36 A-turns/m
From Ampere’s law, Hl c = NI
lc = 2×(8”+6”) = 28” = 0.71m
Hl c (36 )( 0 .71)
I =
=
= 0 .13 A
N
200
Electromechanics
Magnetics and Energy Conversion
2-74
Example: RMS Winding Current
From Figure 1.12, at Bmax = 1.5T, Pa = 1.5
VA/kg
Core volume:
Vc = (4in 2 )(0.94)(28in) = 105.5 in 3 = 1.7 × 10 −3 m 3
Core mass:
M = Vc ρ c = (1.7 × 10 −3 m 3 )(7650
kg
) = 13.2 kg
3
m
Core VoltAmperes: Pc = 1.5
Current: I RMS =
Electromechanics
VA
× 13.2 kg = 19.8 VA
kg
19.7
VA
=
= 0.10 A
E RMS ⎛ 274 ⎞
⎜
⎟
2
⎝
⎠
Magnetics and Energy Conversion
2-75
Example: Core Loss
From Figure 1.14, core loss density = 1.5
W/kg at Bmax = 1.5 Tesla. Total core loss is:
W
Pc = M × 1.5
= (13.2)(1.5) = 20W
kg
Electromechanics
Magnetics and Energy Conversion
2-76
Permanent Magnets
• “Soft” magnetic materials such as magnetic steel can
behave as very weak permanent magnets
• Permanent magnets, or “hard” magnetic materials, have a
high coercive force Hc and can produce significant flux in
an airgap; they also have a “wide” hysteresis loop
Reference: E. Furlani, Permanent Magnet and Electromechanical Devices, Academic Press, 2001, pp. 39
Electromechanics
Magnetics and Energy Conversion
2-77
Brief History of Permanent Magnets
• c. 1000 BC: Chinese compasses using lodestone
– Later used to cross the Gobi desert
Reference: K. Overshott, “Magnetism: it is permanent,” IEE Proceedings-A, vol. 138, no. 1, Jan. 1991, pp. 22-31
Electromechanics
Magnetics and Energy Conversion
2-78
Brief History of Permanent Magnets
Reference: R. Parker, Advances in Permanent Magnetism, John Wiley, 1990, pp. 3
Electromechanics
Magnetics and Energy Conversion
2-79
Brief History of Permanent Magnets (cont.)
• c. 200 BC: Lodestone (magnetite) known to the Greeks
– Touching iron needles to magnetite magnetized them
• 1200 AD: French troubadour de Provins describes use of a
primitive compass to magnetize needles
• 1600: William Gilbert publishes first journal article on
permanent magnets
• 1819: Oersted reports that an electric current moves compass
needle
References:
1. K. Overshott, “Magnetism: it is permanent,” IEE Proceedings-A, vol. 138, no. 1, Jan. 1991, pp. 22-31
2. R. Petrie, “Permanent Magnet Material from Loadstone to Rare Earth Cobalt,” Proc. 1995 Electronics Insulation and
Electrical Manufacturing and Coil Winding Conf., pp. 63-64
3. Rollin Parker, Advances in Permanent Magnetism, John Wiley, 1990
4. E. Hoppe, “Geshichte des Physik,” Vieweg, Braunshweig, 1926, pp. 339
5. W. Gilbert, “De Magnete 1600,” translation by S. P. Thompson, 1900, republished by Basic Books, Inc., New York,
1958
Electromechanics
Magnetics and Energy Conversion
2-80
Brief History of Permanent Magnets (cont.)
• c. 1825: Sturgeon invents
the electromagnet,
resulting in a way to
artificially magnetize
materials
• 7-ounce magnet was able
to lift 9 pounds
References:
1. W. Sturgeon, Mem. Manchester Lit. Phil. Soc., 1846, vol. 7, pp. 625
2. Britannica Online
Electromechanics
Magnetics and Energy Conversion
2-81
Brief History of Permanent Magnets (cont.)
• c. 1830: Joseph Henry
(U.S.) constructs
electromagnets
Joseph Henry
Photo reference: Smithsonian Institute archives
Electromechanics
Magnetics and Energy Conversion
2-82
Brief History of Permanent Magnets (cont.)
• 1917: Cobalt magnet steels developed by Honda and
Takagi in Japan
• 1940: Alnico --- first “modern” material still in common use
– Good for high temperatures
• 1960: SmCo (samarium cobalt) rare earth magnets
– Good thermal stability
• 1983: GE and Sumitomo develop neodymium iron boron
(NdFeB) rare earth magnet
– Highest energy product, but limited temperature range
References:
1. K. Overshott, “Magnetism: it is permanent,” IEE Proceedings-A, vol. 138, no. 1, Jan. 1991, pp. 22-31
2. R. Petrie, “Permanent Magnet Material from Loadstone to Rare Earth Cobalt,” Proc. 1995 Electronics Insulation and
Electrical Manufacturing and Coil Winding Conf., pp. 63-64
Electromechanics
Magnetics and Energy Conversion
2-83
Magnetizing Permanent Magnets
• Material is placed inside magnetizing fixture
• Magnetizing coil is energized with a current producing
sufficient field to magnetize the PM material
Reference: E. Furlani, Permanent Magnet and Electromechanical Devices, Academic Press, 2001, pp. 57
Electromechanics
Magnetics and Energy Conversion
2-84
Pictorial View of Magnetization Process
Reference: R. Parker, Advances in Permanent Magnetism, John Wiley, 1990, pp. 49
Electromechanics
Magnetics and Energy Conversion
2-85
Permanent Magnets
• External effects of permanent magnets can be modeled as
surface current
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 7
Electromechanics
Magnetics and Energy Conversion
2-86
Permanent Magnets
• After magnetization, M has values of either +Msat or -Msat
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 14-15
Electromechanics
Magnetics and Energy Conversion
2-87
Permanent Magnets
• By a constitutive relationship, B = μo(H+M)
• Since M has values of either +Msat or -Msat, it follows that
the slope of the BH curve for the permanent magnet is μo
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 15, 23
Electromechanics
Magnetics and Energy Conversion
2-88
Demagnetization Curves of Ceramic 8
• Typical sintered ceramic magnet
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 62
Electromechanics
Magnetics and Energy Conversion
2-89
Demagnetization Curves for NdFeB
• Strong neodymium-iron-boron
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 74
Electromechanics
Magnetics and Energy Conversion
2-90
Permanent Magnets vs. Steel
• Note that PM has much higher coercive force
Permanent magnet: Alnico 5
Electromechanics
M-5 steel
Magnetics and Energy Conversion
2-91
Lines of Force
• Iron filings follow magnetic field lines
Electromechanics
Magnetics and Energy Conversion
2-92
Cast Alnico
Reference: R. Parker, Advances in Permanent Magnetism, John Wiley, 1990, pp. 65
Electromechanics
Magnetics and Energy Conversion
2-93
Example: PM in Magnetic Circuits
• Fitzgerald, Example 1.9
g = 0.2 cm
lm = 1.0 cm
Am = Ag = 4 cm2
Find flux in airgap Bg
for magnetic
materials
(a) Alnico 5
(b) M-5 steel
Electromechanics
Magnetics and Energy Conversion
2-94
Example: Solution with Alnico PM
NI = 0, so by Ampere’s law:
H m lm + H g g = 0
⎛l
Solve for Hg: H g = − H m ⎜⎜ m
⎝g
⎞
⎟⎟
⎠
Continuity of flux (Gauss’ law):
⎛A ⎞
Ag B g = Am l m ⇒ B g = B m ⎜ m ⎟
⎜A ⎟
⎝ g ⎠
Next, solve for Bm as a function of Hm:
⎛ Ag ⎞
⎛ Ag ⎞
⎜
⎟
⎟⎟
Bm = B g ⎜
= μ o H g ⎜⎜
⎟
⎝ Am ⎠
⎝ Am ⎠
l ⎞⎛ A g ⎞
⎛
⎟⎟ = −6.28 × 10 − 6 H m
⇒ B m = μ o ⎜⎜ − H m m ⎟⎟⎜⎜
g ⎠⎝ Am ⎠
⎝
Plot this load line on Alnico BH curve
Electromechanics
Magnetics and Energy Conversion
2-95
Example: Solution with Alnico
• Result: Bg = 0.3 Tesla
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 89
Electromechanics
Magnetics and Energy Conversion
2-96
Example: Load Line Solution with M-5 Steel
• Use same load line; Bg = 0.38 Gauss (much lower than
with Alnico)
• Note: Earth’s magnetic field ~ 0.5 Gauss
Electromechanics
Magnetics and Energy Conversion
2-97
Some Common Permanent Magnet Materials
• Other tradeoffs not shown here include: mechanical
strength, temperature effects, etc.
Electromechanics
Magnetics and Energy Conversion
2-98
Typical NdFeB B-H Curve
• Neodymium-iron-boron (NdFeB) is the highest strength
permanent magnet material in common use today
• Good material for applications with temperature less than
approximately 80 - 150C
• Cost per pound has reduced greatly in the past few years
• B/H curve below for “grade 35” or 35 MGOe material
Bm, Tesla
Br
1.2
(BH)max
0.6
Hc
Hm, kA/m
-915
Electromechanics
-457.5
Magnetics and Energy Conversion
2-99
NdFeB B-H Curves for Different Grades
Reference: Dexter Magnetics, Inc. http://www.dextermag.com
Electromechanics
Magnetics and Energy Conversion
2-100
Maximum Energy Product
• BH has units of Joules per unit volume
Electromechanics
Magnetics and Energy Conversion
2-101
Why is maximum energy product important?
Maximum Energy
Product
(1) B
g
⎛ Am
= Bm ⎜
⎜ A
⎝ g
⎞
⎟
⎟
⎠
H mlm
= −1
Hgg
Let’s find Bg2
(2)
⎛
Am ⎞⎟
⎜
× (μ o H g )
B = Bm
⎜
⎟
A
g ⎠
⎝
⎛
A ⎞⎛ H l ⎞
= −⎜ Bm m ⎟⎜⎜ μ o m m ⎟⎟
⎜
Ag ⎟⎠⎝
g ⎠
⎝
⎛ Vol mag ⎞
⎟(− Bm H m )
= μo ⎜
⎜ Vol ⎟
gap ⎠
⎝
2
g
Solve for magnet volume Volmag
Vol mag
⎛ Vol gap
= B ⎜⎜
⎝ − Bm H m
2
g
⎞
⎟⎟
⎠
To use minimum volume of magnet for a
given Bg, operate magnet at (BH)max point
Electromechanics
Magnetics and Energy Conversion
2-102
Progress in PM Specs
• One figure of merit is (BH)max product
Reference: J. Evetts, Concise Encyclopedia of Magnetic and Superconducting Materials, Pergamon Press, Oxford, 1992
Electromechanics
Magnetics and Energy Conversion
2-103
Progress in PM Specs
Reference: K. Overshott, “Magnetism: it is permanent,” IEE Proceedings-A, vol. 138, no. 1, Jan. 1991, pp. 22-31
Electromechanics
Magnetics and Energy Conversion
2-104
Applications for Permanent Magnets
• Disk drives
• Speakers
• Motors
– Rotary motors (Toyota Prius)
– Linear motors (Maglev, people moving)
• Refrigerator magnets
• Proximity sensors and switches
• Compasses
• Magnetic bearings and magnetic suspensions (Maglev)
• Water filtration
• Plasma fusion research, NMR
• Eddy current brakes (ECBs)
• Etc.
References:
1. R. Parker, Advances in Permanent Magnetism, John Wiley, 1990
2. P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994
Electromechanics
Magnetics and Energy Conversion
2-105
Example: Maximum Energy Product
• Fitzgerald, Example 1.10: Find magnet dimensions for
desired airgap flux density Bg = 0.8 Tesla
At maximum (BH), Bm=1.0 T and Hm = -40 kA/m
⎛ Bg ⎞
0 .8 ⎞
2
⎟⎟ = (2cm 2 )⎛⎜
Am = Ag ⎜⎜
⎟ = 1.6cm
⎝ 1 .0 ⎠
⎝ Bm ⎠
⎛ Hg
l m = − g ⎜⎜
⎝ Hm
⎞
⎛ Bg
⎟⎟ = − g ⎜⎜
⎠
⎝ μo H m
⎞
⎞
⎛
0.8
⎟⎟ = (−0.2cm)⎜⎜
⎟⎟ = 3.18cm
−7
π
(
4
10
)(
40000
)
×
−
⎠
⎝
⎠
So, magnet is 3.18 cm long and 1.6 cm2 in area
Electromechanics
Magnetics and Energy Conversion
2-106
Open-Circuited Permanent Magnet
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 89
Electromechanics
Magnetics and Energy Conversion
2-107
Open Circuited Permanent Magnet --- FEA
Electromechanics
Magnetics and Energy Conversion
2-108
Short-Circuited Permanent Magnet
• Find B inside core, ignoring any leakage and assuming
infinite permeability in core
Electromechanics
Magnetics and Energy Conversion
2-109
Short Circuited Permanent Magnet
• For infinite permeability, load line is vertical
• Intersection of load lines occurs at B ≈ Br
Reference: P. Campbell, Permanent Magnet Materials and their Applications, Cambridge University Press, 1994, pp. 89
Electromechanics
Magnetics and Energy Conversion
2-110
Short Circuited Permanent Magnet --- FEA
Electromechanics
Magnetics and Energy Conversion
2-111
Circuit Modeling of
Permanent Magnets
Reference: E. P. Furlani, Permanent Magnet and Electromechanical Devices, Academic Press, 2001
Electromechanics
Magnetics and Energy Conversion
2-112
Circuit Modeling of Permanent Magnets
Electromechanics
Magnetics and Energy Conversion
2-113
Circuit Modeling of Permanent Magnets
Electromechanics
Magnetics and Energy Conversion
2-114
Circuit Modeling of Permanent Magnets
Electromechanics
Magnetics and Energy Conversion
2-115
Example: Circuit Modeling of Permanent
Magnets
Electromechanics
Magnetics and Energy Conversion
2-116
Example: Circuit Modeling
of Permanent Magnets
Electromechanics
Magnetics and Energy Conversion
2-117
Example: Circuit Modeling of Permanent
Magnets---FEA
Electromechanics
Magnetics and Energy Conversion
2-118
Example: Magnetic Circuit With Steel
• Estimate airgap field Bg assuming grade 37 NdFeB
• Airgap g, magnet thickness tm
Electromechanics
Magnetics and Energy Conversion
2-119
Example: Magnetic Circuit With Steel
• This analysis also ignores leakage
Electromechanics
Magnetics and Energy Conversion
2-120
Example:
Magnetic Circuit
With Steel --Operating Point
vs. Magnet
Thickness tm
Electromechanics
Magnetics and Energy Conversion
2-121
Example: FEA with Magnet Thickness tm = g/2
Electromechanics
Magnetics and Energy Conversion
2-122
Example: FEA with Magnet Thickness tm = g
Electromechanics
Magnetics and Energy Conversion
2-123
Example: FEA with Magnet Thickness tm = 2g
Electromechanics
Magnetics and Energy Conversion
2-124
Example: Comparison of Different Magnet
Thicknesses
1.00E+00
8.00E-01
6.00E-01
4.00E-01
2.00E-01
0.00E+00
0.00E+001.00E+002.00E+003.00E+004.00E+005.00E+006.00E+007.00E+008.00E+009.00E+00
-2.00E-01
-4.00E-01
-6.00E-01
-8.00E-01
-1.00E+00
Electromechanics
Magnetics and Energy Conversion
2-125
PM and a Winding
• Many motors have permanent magnets, steel and windings
Analysis of PM in closed core,
with excitation:
H m l m = NI
∴Hm =
Electromechanics
NI
lm
Magnetics and Energy Conversion
2-126
PM and a Winding --- Load Line
• Note that demagnetization can occur if current is
sufficiently high
Electromechanics
Magnetics and Energy Conversion
2-127
Another Example --- Excitation and Airgap
(1) Ampere’s law:
H m l m + H g g = NI
(2) Constitutive law: B g = μ o H g
(3) Gauss’ law: Bm Am = B g Ag
Solve for Bm-Hm load line:
⎛A l
Bm = − μ o ⎜⎜ m m
⎝ Am g
Electromechanics
⎞⎛
NI ⎞
⎟⎟
⎟⎟⎜⎜ H m −
lm ⎠
⎠⎝
Magnetics and Energy Conversion
2-128
Another Example --- Excitation and Airgap --Load Line
Electromechanics
Magnetics and Energy Conversion
2-129
Interesting Calculation Tool
Reference: www.magnetsales.com
Electromechanics
Magnetics and Energy Conversion
2-130
What Can You
Buy?
Reference: www.dextermag.com
Electromechanics
Magnetics and Energy Conversion
2-131
What Can You
Buy?
Reference: www.dextermag.com
Electromechanics
Magnetics and Energy Conversion
2-132
What Can You
Buy?
Reference: www.magnetsales.com
Electromechanics
Magnetics and Energy Conversion
2-133
Magnetization Patterns
Reference: www.magnetsales.com
Electromechanics
Magnetics and Energy Conversion
2-134
Comparison
Reference: www.magnetsales.com
Electromechanics
Magnetics and Energy Conversion
2-135
Comparison of Maximum Operating
Temperatures
Reference: http://www.electronenergy.com/media/Magnetics%202005.pdf
Electromechanics
Magnetics and Energy Conversion
2-136
Comparison of Maximum Operating
Temperatures
Reference: http://www.electronenergy.com/media/Magnetics%202005.pdf
Electromechanics
Magnetics and Energy Conversion
2-137
Magnet Comparisons
Reference: www.magnetsales.com
Electromechanics
Magnetics and Energy Conversion
2-138
PM Online Resources
•
•
•
•
•
http://members.aol.com/marctt/
www.dextermag.com
www.magnetsales.com
http://www.grouparnold.com/
Magnetic Materials Producers Association (MMPA
standard)
Electromechanics
Magnetics and Energy Conversion
2-139
Some Very Brief Comments on Superconductors
• Superconductors have zero resistance if the temperature is
low enough, the field acting on the superconductor is low
enough, and the current through the superconductor is low
enough
• Superconductors are classified as “low temperature” (NbTi,
NbSn) or “high temperature” (YBCO, BSCCO)
• Low-Tc superconductors are usually chilled with liquid helium
(4.2K)
• High-Tc superconductors are usually used in the 20K-77K
range
Reference: Y. Iwasa “Case Studies in Superconducting Magnets,” Plenum Press, 1994
Electromechanics
Magnetics and Energy Conversion
2-140
Some Data on Low-Tc Material
• Shown for niobium titanium
• This type of superconductor is used in the Japanese MLX
500 km/hr Maglev
Electromechanics
Magnetics and Energy Conversion
2-141
Some Data on High-Tc Material
• Some superconductors are anisotropic; i.e. superconducting
tapes
Reference: American Superconductor, www.amsuper.com
Electromechanics
Magnetics and Energy Conversion
2-142
Quotes
It is well to observe the force and virtue and consequence of
discoveries, and these are to be seen nowhere more
conspicuously than in printing, gunpowder, and the magnet.
--- Sir Francis Bacon
The mystery of magnetism, explain that to me! No greater
mystery, except love and hate.
---John Wolfgang von Goethe
Electromechanics
Magnetics and Energy Conversion
2-143
Transformers --- Overview
• Selected history
• Types of transformers
• Voltages and currents
• Equivalent circuits
• Voltage and current transformers
• Per-unit system
Electromechanics
Magnetics and Energy Conversion
2-144
Selected History
• 1831 --- Transformer action
demonstrated by Michael
Faraday
• 1880s: modern transformer
invented
Reference: J. W. Coltman, “The Transformer (historical overview),” IEEE
Industry Applications Magazine, vol. 8, no. 1, Jan.-Feb. 2002, pp. 8-15
Electromechanics
Magnetics and Energy Conversion
2-145
Early Transformer (Stanley, c. 1880)
William Stanley
Electromechanics
Magnetics and Energy Conversion
2-146
Faraday’s Law
• A changing magnetic flux impinging on
a conductor creates an electric field
and hence a current (eddy current)
d
∫C E ⋅ dl = − dt ∫S B ⋅ dA
• The electric field integrated around a
closed contour equals the net timevarying magnetic flux density flowing
through the surface bound by the
contour
• In a conductor, this electric field creates
a current by:
J = σE
• Induction motors, brakes, etc.
Electromechanics
Magnetics and Energy Conversion
2-147
Magnetic Circuit with Two Windings
• Note that flux is the sum of flux due to i1 and that due to i2
Electromechanics
Magnetics and Energy Conversion
2-148
Types of Transformers
• There are many different types and power ratings of
transformers: single phase and multi-phase, signal
transformers, current transformers, etc.
Electromechanics
Magnetics and Energy Conversion
2-149
Instrument Transformer
• Instrument transformers (voltage and current) provide line
current and line voltage information to protective relays
and control systems
• Current transformer shown below
Reference: L. Faulkenberry and W. Coffer, Electrical Power Distribution and Transmission, Prentice Hall, 1996, pp. 131
Electromechanics
Magnetics and Energy Conversion
2-150
200A Current Transformer (CT)
Reference: http://rocky.digikey.com/WebLib/Amveco-Talema/Web%20Data/AC1200.pdf
Electromechanics
Magnetics and Energy Conversion
2-151
Voltage Instrument Transformer
Reference: http://www.geindustrial.com/products/brochures/ITI.pdf
Electromechanics
Magnetics and Energy Conversion
2-152
Audio Transformer
Reference: http://rocky.digikey.com/WebLib/Hammond/Web%20Data/560,%20800-844,%20850%20Series.pdf
Electromechanics
Magnetics and Energy Conversion
2-153
Pulse Transformer
• Used for triggering SCRs, etc. where isolation is needed
Reference: http://rocky.digikey.com/WebLib/Tamura-Microtran/Web%20Data/STT-107.pdf
Electromechanics
Magnetics and Energy Conversion
2-154
Power Distribution Transformer
• Provides voltage for the customer
• Typical voltages are 2.3-34.5kV primary, and 480Y/277V or
208Y/120V 3-phase or 240/120V single phase
• Pole-top transformers typically 15-100 kVA
Reference: http://www.geindustrial.com/products/brochures/DEA-271-English.pdf
Electromechanics
Magnetics and Energy Conversion
2-155
E Core Transformer
Electromechanics
Magnetics and Energy Conversion
2-156
Offline Flyback Power Supply
P. Maige, “A universal power supply integrated circuit for TV and monitor applications,” IEEE Transactions on Consumer
Electronics, vol. 36, no. 1, Feb. 1990, pp. 10-17
Electromechanics
Magnetics and Energy Conversion
2-157
Transcutaneous Energy Transmission
H. Matsuki, Y. Yamakata, N. Chubachi, S.-I. Nitta and H. Hashimoto, “Transcutaneous DC-DC converter for totally
implantable artificial heart using synchronous rectifier,” IEEE Transactions on Magnetics, vol. 32 , no. 5 , Sept. 1996, pp.
5118 - 5120
Electromechanics
Magnetics and Energy Conversion
2-158
Power Transformer
Reference: J. W. Coltman, “The Transformer (historical overview),” IEEE Industry Applications Magazine, vol. 8, no. 1, Jan.Feb. 2002, pp. 8-15
Electromechanics
Magnetics and Energy Conversion
2-159
Superconducting Transformer
Reference: W. Hassenzahl et. al., “Electric Power Applications of Superconductivity,” Proceedings of the IEEE, vol. 92, no.
10, October 2004, pp. 1655-1674
Electromechanics
Magnetics and Energy Conversion
2-160
Transformer Under No-Load Condition
By Faraday’s law:
dλ1
dΦ
e1 =
=N
dt
dt
= ωNΦ max cos(ωt )
RMS value of e1:
2πfNΦ max
E1,rms =
= 2πfNΦ max
2
If resistive drop in winding is
negligible:
E1,rms
Φ max =
2πfN
Electromechanics
Magnetics and Energy Conversion
2-161
No-Load Phasor Diagram
• Winding current has harmonics,
and fundamental is generally out
of phase with respect to flux
• In-phase component is from
core losses
Pc = E1 Iϕ cos θ c
• Magnetizing current is 90
degrees out of phase
Electromechanics
Magnetics and Energy Conversion
2-162
Example: Transformer Calculations
• Fitzgerald, Example 2.1. In Example 1.8, the core loss
and VA at Bmax = 1.5T and 60 Hz were found to be: Pc =
16W and VI = 20 VA with induced voltage 194V. Find
power factor, core loss current Ic and magnetizing current
Im.
Electromechanics
Magnetics and Energy Conversion
2-163
Example: Transformer Calculations --- Solution
Power factor:
16
PF = cos(θ c ) =
= 0 .8
20
Exciting current
VA 20
Iϕ =
=
= 0 .1 A
V
194
Core loss component
16
Ic =
= 0.082 A
194
Magnetizing component
I m = I ϕ sin θ c = 0.060 A
Electromechanics
Magnetics and Energy Conversion
2-164
Ideal Transformer with Load
Ideal transformer:
dΦ
v1 = N 1
dt
dΦ
v2 = N 2
dt
v2 N 2
=
v1 N 1
By Ampere’s law:
i1 N 2
N 1i1 − N 2 i2 = 0 ⇒ =
i2 N1
(Also, by power balance,
v1i1 = v 2 i2 )
Electromechanics
Magnetics and Energy Conversion
2-165
Impedance Transformation
N1 ˆ
ˆ
V1 =
V2
N2
ˆ
⎛ N 2 ⎞⎛ N 2
N
N
V
2
2
2
ˆI =
ˆI =
⎟⎟⎜⎜
= ⎜⎜
2
1
N1
N 1 Z 2 ⎝ N 1 ⎠⎝ N 1
⎞ˆ⎛ 1
⎟⎟V1 ⎜⎜
⎠ ⎝ Z2
⎞
⎟⎟
⎠
Therefore, impedance at input terminals is:
2
ˆ
⎛ N1 ⎞
V1
⎟⎟
= Z 2 ⎜⎜
Iˆ1
⎝ N2 ⎠
Electromechanics
Magnetics and Energy Conversion
2-166
Equivalent Circuits
• These 3 circuits have the same impedance as seen from
the a-b terminals
Electromechanics
Magnetics and Energy Conversion
2-167
Example: Use of Equivalent Circuits
• Fitzgerald, Example 2.2
• (a) Draw equivalent circuit with series impedance
referred to primary
• (b) For a primary voltage of 120VAC and a short at the
output, find the primary current and the short circuit
current at the output
Electromechanics
Magnetics and Energy Conversion
2-168
Example: Impedance Transformation
2
⎛ N1 ⎞
⎟⎟ (R2 + jX 2 )
R + jX = ⎜⎜
⎝ N2 ⎠
= 25 + j100
'
2
Electromechanics
'
2
Magnetics and Energy Conversion
2-169
Example: Input Current with Output Short Circuit
Iˆ1 =
Vˆ1
120 ⎛ 25 − j100 ⎞
⎟⎟
⎜⎜
=
'
'
R2 + jX 2 25 + j100 ⎝ 25 − j100 ⎠
Iˆ1 = 0.28 − j (1.13)
I1,rms = 0.282 + 1.132 = 1.16 A
Iˆ2 = 5 Iˆ1 = 1.4 − j (5.65)
I 2,rms = 1.4 2 + 5.652 = 5.8 A
Electromechanics
Magnetics and Energy Conversion
2-170
Non-Ideal Effects in Transformers --- Magnetizing
Inductance
• A real-world transformer doesn’t pass DC
• From either set of terminals, the impedance looks like an
inductor if the other set of terminals is open-circuited
• Can model this as an ideal transformer with a magnetizing
inductance added.
• The magnetizing current im produces the mutual flux which
couples to the secondary
Electromechanics
Magnetics and Energy Conversion
2-171
Mutual and Leakage Flux
• Not all of the flux created by winding #1 links with winding #2.
Electromechanics
Magnetics and Energy Conversion
2-172
Mutual and Leakage Flux
Electromechanics
Magnetics and Energy Conversion
2-173
Non-Ideal Effects in Transformers --- Leakage
• Not all of the flux created by winding #1 links with winding #2.
• Therefore, real-world voltage transformation is not exactly
equal to the turns ratio, due to the voltage drops on Lk1 and
Lk2
Electromechanics
Magnetics and Energy Conversion
2-174
Transformer Equivalent Circuits
Electromechanics
Magnetics and Energy Conversion
2-175
Example: Use of Equivalent Circuits
• Fitzgerald, Example 2.3: A 50-kVA 2400:240V 60 Hz
distribution transformer has a leakage impedance of
0.72+j0.92Ω in the high voltage winding and 0.0070 +
j0.0090 Ω in the low-voltage winding. The impedance Zϕ
of the shunt branch (equal to Rc +jXm in parallel) is 6.32 +
j43.7 Ω when viewed from the low voltage side.
Draw the equivalent circuits referred to the high voltage
side and the low voltage sides.
SOLUTION:
Note that N1:N2 is 1:10, so impedances step up and
down by 100
Electromechanics
Magnetics and Energy Conversion
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Example: Solution
--- Leakage impedance of
0.72+j0.92Ω in the high
voltage winding and
0.0070 + j0.0090 Ω in the
low-voltage winding.
--- The impedance Zϕ of
the shunt branch (equal
to Rc +jXm in parallel) is
6.32 + j43.7 Ω when
viewed from the low
voltage side.
Electromechanics
Magnetics and Energy Conversion
2-177
Approximate Transformer Equivalent Circuits
• “Cantilever circuits”
• Ignoring voltage drop in primary or secondary leakage
impedances
Electromechanics
Magnetics and Energy Conversion
2-178
Approximate Transformer Equivalent Circuits
• Circuit if we ignore the magnetizing inductance and core
resistance
• Circuit if we further ignore the winding resistance
Electromechanics
Magnetics and Energy Conversion
2-179
Example: Use of Cantilever Circuit
• Fitzgerald, Example 2.4
Using T-model:
⎛ Zϕ
ˆ
V2 = (2400 )⎜
⎜Z +Z
ϕ
⎝ l1
⎞⎛ 1 ⎞
⎟⎜ ⎟ = 239.9 + j 0.0315
⎟⎝ 10 ⎠
⎠
Using cantilever model:
⎛1⎞
Vˆ2 = (2400 )⎜ ⎟ = 240
⎝ 10 ⎠
Electromechanics
Magnetics and Energy Conversion
2-180
Example: Find V2
• Fitzgerald, Example 2.5
Electromechanics
Magnetics and Energy Conversion
2-181
Example: Find V2
From node equations:
Vˆs = Vˆ2 + IˆL R + jXIˆL
Electromechanics
Magnetics and Energy Conversion
2-182
Example: Find V2
-We need length of vector Oa (which is V2)
-We know length of vector Oc (which is 2400V)
ab = IR cos θ + IX sin θ = (20.8)(1.72)(0.8) + (20.8)(3.42)(0.6) = 71.4
bc = IX cos θ − IR sin θ = 35.5
Solve for V2:
(V2 + ab) 2 + (bc ) 2 = Vs2 ⇒ V2 = 233V
Electromechanics
Magnetics and Energy Conversion
2-183
Transformer Testing to Determine Parameters
• By doing various tests on a transformer, we can
determine the equivalent circuit parameters
• Testing includes open-circuit and short-circuit testing
Electromechanics
Magnetics and Energy Conversion
2-184
Short-Circuit Test
Electromechanics
Magnetics and Energy Conversion
2-185
Short-Circuit Testing
Normally:
Lm >> Lk1, Lk2 and Rc >> Rw1, Rw2
Using the simplified circuit, we can
approximate:
V
Z eq ≈ TEST
I TEST
P
RSC = RW 1 + RW' 2 = 2SC
I TEST
X SC ≈
Electromechanics
Z eq
2
2
− RSC
Magnetics and Energy Conversion
2-186
Open-Circuit Testing
Electromechanics
Magnetics and Energy Conversion
2-187
Open-Circuit Testing
• There is no secondary current
2
VTEST
Rc ≈
POC
Xm
Electromechanics
VTEST
≈
I TEST
Magnetics and Energy Conversion
2-188
Autotransformer
Conventional transformer
Autotransformer redrawn
Connection as autotransformer
Electromechanics
Magnetics and Energy Conversion
2-189
Staco Autotransformer
Reference: Staco
Electromechanics
Magnetics and Energy Conversion
2-190
Autotransformer Drawing
Reference: Staco
Electromechanics
Magnetics and Energy Conversion
2-191
Autotransformer Brush
Electromechanics
Magnetics and Energy Conversion
2-192
Damaged Autotransformer
Electromechanics
Magnetics and Energy Conversion
2-193
Example: Autotransformer
• Fitzgerald, Example 2.7
• 2400:240V 50-kVA transformer is connected as an
autotransformer with ab being the 240V winding and bc is
the 2400V winding.
(a) Compute the kVA rating
(b) Find currents at rated power
Electromechanics
Magnetics and Energy Conversion
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Example: Autotransformer Example --- Solution
For 50 kVA, rating of 240V winding is:
50000 / 240 = 208 A
The autotransformer VA rating is:
V H I H = (2.64kV )(208 A) = 549 kVA
Rated current at low-voltage winding:
⎛ 2640 ⎞
IL = IH ⎜
⎟ = 229 A
⎝ 2400 ⎠
Electromechanics
Magnetics and Energy Conversion
2-195
Some Comments on Autotransformers
• Autotransformers differ from isolation transformers in that
there is no isolation between primary and secondary
• However, this lack of isolation allows some of the
transferred power to be conducted from primary to
secondary instead of magnetic induction
• Autotransformers in general require less core material
per kVA rating
• Autotransformers used where lack of isolation doesn’t
pose a safety issue
Electromechanics
Magnetics and Energy Conversion
2-196
Example: Magnetic Circuit Problem
• Fitzgerald, Problem 2.2
• A magnetic circuit with a cross-sectional area of 15 cm2
is to be operated from a 120V RMS supply. Calculate the
number of turns required to achieve a peak magnetic flux
density of 1.8 Tesla in the core
Electromechanics
Magnetics and Energy Conversion
2-197
Flux is: Φ = Φ max sin(ωt )
Example:
Magnetic
Circuit Problem
--- Solution
The time rate of change of flux is:
dΦ
= ωΦ max cos(ωt )
dt
The time rate of change of flux linkage is:
dλ
dΦ
=N
= NωΦ max cos(ωt ) = V
dt
dt
Let’s relate flux density to flux:
Φ max
= Bmax
A
So, we can solve for N
N=
2V
ωBmax A
=
( 2 )(120)
= 166.7
( 2π × 60)(1.8)(1.5 × 10 −3 )
Round up to N = 167
Electromechanics
Magnetics and Energy Conversion
2-198
Example: Transformer Problem
• Fitzgerald, Problem 2.4
• A 100-Ohm resistor is connected to the secondary of an
ideal transformer with a turns ratio of 1:4 (primary to
secondary). A 10V RMS, 1-kHz voltage source is
connected to the primary. Calculate the primary current
and the voltage across the 100-Ohm resistor
Electromechanics
Magnetics and Energy Conversion
2-199
Example: Transformer Problem --- Solution
We know that this is a step-up transformer, so
V1 = 10V and V2 = 40V.
We next find the secondary current I2
40V
I2 =
= 0.4 A
100Ω
The voltage steps up, so the current steps down;
hence
I1 = 4I2 = 1.6A
Electromechanics
Magnetics and Energy Conversion
2-200
3-Phase Connections of Transformers
Electromechanics
Magnetics and Energy Conversion
2-201
Example: Finding 3-Phase Currents
• Problem: A three phase, 208V (line-line) Y connected
load has:
– Zan = 3 + j4
– Zbn = 5
– Zcn = -5j
• Find
– (a) Phase voltages
– (b) Line and phase currents
– (c) Neutral currents
Electromechanics
Magnetics and Energy Conversion
2-202
Example: Finding 3-Phase Currents --- Solution
Line-neutral voltages are found by taking the lineline voltage and dividing by 3
208
= 120
VL − N =
3
Electromechanics
Magnetics and Energy Conversion
2-203
Example: 3-Phase Currents --- Solution
Line (also called phase) currents are found by taking the
line-line voltages and dividing by impedance
120∠0 o
= 24∠ − 53.13o
Ia =
(3 + 4 j )
120∠120 o
= 24∠120 o
Ib =
(5)
120∠240 o
= 24∠ − 30 o
Ic =
(−5 j )
Electromechanics
Magnetics and Energy Conversion
2-204
Example: 3-Phase Currents --- Solution
• Note that loads are unbalanced so there is a net neutral
current
I n = I a + I b + I c = 25.4∠155.9
Electromechanics
o
Magnetics and Energy Conversion
2-205
480V Y System
• Line-line = 480V; line-neutral = 277V
Reference: Ralph Fehr, Industrial Power Distribution, Prentice Hall, 2002
Electromechanics
Magnetics and Energy Conversion
2-206
Delta System
• Line-line = 480V
Reference: Ralph Fehr, Industrial Power Distribution, Prentice Hall, 2002
Electromechanics
Magnetics and Energy Conversion
2-207
Instrument Transformer
• Used for protection, relaying, voltage or current monitoring,
etc.
Electromechanics
Magnetics and Energy Conversion
2-208
Per-Unit System
• Computations in electric machines and transformers are
often done using the “per-unit” system
• Actual circuit quantities (Watts, VArs, etc.) are scaled to
the per-unit system
• This method allows removal of transformers from
diagrams
• To convert to per-unit, 4 base quantities are established
– Base power VAbase
– Base voltage Vbase
– Base current Ibase
– Base impedance Zbase
Electromechanics
Magnetics and Energy Conversion
2-209
Example: Per-Unit System
• Example: A system has Zbase = 10 Ω and Vbase = 400V.
Find base VA and Ibase
• Solution:
– Ibase = Vbase/Zbase = 400/10 = 40A
– (VA)base = VbaseIbase = (400)(40) = 16 kVA
Electromechanics
Magnetics and Energy Conversion
2-210
Example: Per-Unit System
• A 5 kVA, 400/200V transformer has 2 Ω reactance
referred to the 200V side. Express the transformer
reactance in p.u.
• Solution:
– (VA)base = 5000
– Vbase = 200
– Ibase = (VA)base/Vbase = 5000/200 = 25A
– Zbase =Vbase/Ibase = 200/25 = 8 Ω
– Z = 2.0/8.0 = 0.25 p.u.
Electromechanics
Magnetics and Energy Conversion
2-211
Example: Per-Unit Applied to Transformer
• Fitzgerald, Example 2.12
• Convert this circuit showing a 100 MVA transformer to the
per-unit system
Electromechanics
Magnetics and Energy Conversion
2-212
Example: Low-Voltage Side
(VA) BASE = 100 MVA
V BASE = 7.99 kV
R BASE = X BASE = Z BASE
⎛ (7.99 × 10 3 ) 2
= ⎜⎜
−6
100
10
×
⎝
2
V BASE
=
(VA) BASE
⎞
⎟⎟ = 0.638 Ω
⎠
In per-unit system:
0.76 × 10 −3
RL =
= 0.0012 p.u.
0.638
0.04
XL =
= 0.063 p.u.
0.638
114
Xm =
= 178.7 p.u.
0.638
Electromechanics
Magnetics and Energy Conversion
2-213
Example: High-Voltage Side
(VA) BASE = 100 MVA
VBASE = 79.7 kV
R BASE = X BASE = Z BASE
⎛ (79.7 × 10 3 ) 2
= ⎜⎜
−6
×
100
10
⎝
2
V BASE
=
(VA) BASE
⎞
⎟⎟ = 63.5 Ω
⎠
In per-unit system:
0.085
RH =
= 0.00133 p.u.
63.5
3.75
XH =
= 0.059 p.u..
63.5
Electromechanics
Magnetics and Energy Conversion
2-214
Example: Model
• Turns ratio is 1:1, so we can remove transformer
Electromechanics
Magnetics and Energy Conversion
2-215
Example: Get Rid of 1:1 Transformer
Electromechanics
Magnetics and Energy Conversion
2-216
Energy Conversion --- Overview
• Forces and torques
• Energy balance
• Determining magnetic forces and torques from energy
• Multiply-excited systems
• Forces and torques in systems with permanent magnets
Electromechanics
Magnetics and Energy Conversion
2-217
Right-Hand Rule
• For determining the direction magnetic-field component of
the Lorentz force F=q(v x B) = JxB.
Electromechanics
Magnetics and Energy Conversion
2-218
Example: Single-Coil Rotor
• Fitzgerald, Example 3.1
• Find θ-directed torque as a function of α
Electromechanics
Magnetics and Energy Conversion
2-219
Example: Single-Coil Rotor
For wire #1:
F1θ = − IlBo sin α
For wire #2:
F2θ = − IlBo sin α
Total torque (T = force × distance):
Tθ = −2 IlBo sin αR
What happens if B points left-right
instead
of up-down?
Tθ = −2 IlBo cos αR
Electromechanics
Magnetics and Energy Conversion
2-220
Electromechanical Energy Conversion Device
• This box can be used to model motors, actuators, lift
magnets, etc.
• Note 2 electrical terminals (voltage and current) and 2
mechanical terminals (force ffld and position x)
• The lossless magnetic energy storage system converts
electrical energy to mechanical energy
Electromechanics
Magnetics and Energy Conversion
2-221
Interaction Between Electrical and Mechanical
Terminals
WFLD = stored magnetic energy
In words, the rate of change of magnetic energy equals
the power in minus the mechanical work out
dWFLD
dx
= ei − f fld
dt
dt
By Faraday’s law, e = dλ/dt, so let’s rework:
dWFLD
dx
dλ
=i
− f fld
dt
dt
dt
Multiply through everywhere by dt:
dWFLD = idλ − f fld dx
Electromechanics
Magnetics and Energy Conversion
2-222
Interaction Between Electrical and Mechanical
Terminals
In a lossless system, we can rewrite the energy
balance:
dW ELEC = dWMECH + dW FLD
Differential energy in: dW ELEC = idλ
Differential work out: dWMECH = f fld dx
Change in magnetic energy: dWFLD
Electromechanics
Magnetics and Energy Conversion
2-223
Force-Producing Device
• This solenoid is an example of a force-producing device
Electromechanics
Magnetics and Energy Conversion
2-224
Energy
Thinking about energy, we start out with:
dW FLD = idλ − f fld dx
If magnetic energy storage is lossless, this is
a conservative system and W fld is determined
by state variables λ and x
In a conservative system, the path you take to
do this integration doesn’t matter
Electromechanics
Magnetics and Energy Conversion
2-225
Another Conservative System
• Roller coaster, ignoring friction, the path doesn’t
matter. Speed of both coasters is the same at the
bottom of the hill
Electromechanics
Magnetics and Energy Conversion
2-226
Magnetic Relay
• This illustrates lossless magnetic structure with external
losses due to resistance
Electromechanics
Magnetics and Energy Conversion
2-227
Integration Path for Finding Magnetic Stored
Energy
∫ dW
WFLD (λo , xo ) =
FLD
path 2 a
+
∫ dW
FLD
path 2 b
On path 2a, dλ = 0 and ffld = 0 since zero λ
means zero magnetic force, therefore:
λ
WFLD (λo , xo ) = ∫ i (λ , xo )dλ
0
Electromechanics
Magnetics and Energy Conversion
2-228
Special Case --- Linear System
In the special case of a linear system, the
flux linkage is proportional to current, or λ ∝
i.
λ
WFLD (λ , x) = ∫ i (λ' , x)dλ'
0
λ
2
λ
1
=∫
dλ ' =
L( x)
2 L( x)
0
Electromechanics
λ'
Magnetics and Energy Conversion
2-229
Example: Relay with Movable Plunger
• Fitzgerald, Example 3.2
• Find magnetic stored energy WFLD as a function of x with
I = 10A
Electromechanics
Magnetics and Energy Conversion
2-230
Example: Relay with Movable Plunger
1 λ2
with I constant.
WFLD =
2 L( x)
We know that λ = L( x) I , so
1 L2 ( x)I 2 1
WFLD =
= L( x)I 2
2 L( x)
2
μo N 2
x
I’ll give you that L( x) =
ld (1 − ) ,
2g
d
so magnetic stored energy is:
μo N 2
x 2
WFLD =
ld (1 − )I
4g
d
Electromechanics
Magnetics and Energy Conversion
2-231
Determining Magnetic Force from Stored Energy
• Next, if we go to all
this trouble to find
stored energy, let’s
figure out how to
find forces from the
energy (very
important!)
Remember dW FLD = idλ − f fld dx
Next, remember the “total differential”
from calculus:
∂F
∂F
dF ( x1 , x2 ) =
dx1 +
dx2
∂x1 x =const .
∂x2 x =const .
2
1
Let’s rewrite the stored energy
expression:
∂WFLD
∂WFLD
dλ +
dWFLD =
dx λ
dλ x
From this, we see that
∂WFLD
∂W
and f fld = − FLD
i=
dλ x
∂x
Electromechanics
λ
Magnetics and Energy Conversion
2-232
Determining Magnetic Force from Energy
For linear systems with λ = L(x)I
1 λ2
Energy WFLD =
2 L( x)
f fld
Force:
∂ ⎛ 1 λ2 ⎞
⎟⎟
= − ⎜⎜
∂x ⎝ 2 L( x) ⎠ λ =cons tan t
λ2
dL( x)
= 2
2 L ( x) dx
With λ = L(x)I
I 2 dL( x)
f fld =
2 dx
Electromechanics
Magnetics and Energy Conversion
2-233
Determining Magnetic Force from Energy
• The bottom line: if your system is linear, and if you can
calculate inductance as a function of position, then finding the
force is pretty easy
Electromechanics
Magnetics and Energy Conversion
2-234
Example: Curve Fit for Inductance of Solenoid
with Plunger
• Fitzgerald, Example 3.3
• Assume that the following inductance vs. plunger position
was measured. We then run the solenoid with 0.75A current
Electromechanics
Magnetics and Energy Conversion
2-235
Example: Force as a Function of Position
• We can fit a polynomial to the inductance, and use energy
methods to find the plunger force as a function of position
Electromechanics
Magnetics and Energy Conversion
2-236
Torque in Magnetic Circuit
• Can solve this as before, by analogy
By analogy:
I 2 dL (θ )
T fld =
2 dθ
Electromechanics
Magnetics and Energy Conversion
2-237
Example: Finding Torque
• Fitzgerald, Example 3.4
Assume L(θ) = Lo + L2cos(2θ) with Lo = 10.6 mH and L2 = 2.7 mH.
Find the torque with I = 2A.
Solution:
I 2 dL(θ )
I2
T (θ ) =
= + (− 2 L2 sin( 2θ ) )
2 dθ
2
= −1.08 × 10 − 2 sin( 2θ ) N − m
Electromechanics
Magnetics and Energy Conversion
2-238
Example: Finding Torque
• Fitzgerald, Practice Problem 3.4
Assume L(θ) = Lo + L2cos(2θ) + L4sin(4θ) with Lo = 25.4 mH, L2 = 8.3 mH
and L4 = 1.8 mH. Find the torque with I = 3.5A.
Solution:
T (θ ) = −0.1017 sin( 2θ ) + 0.044 cos( 4θ ) N − m
Electromechanics
Magnetics and Energy Conversion
2-239
Another Example --- Torque vs. Rotor Angle
0.6
0.4
Torque [N-m]
0.2
0
-0.2
-0.4
-0.6
-0.8
Electromechanics
0
50
100
150
200
250
Theta [deg]
300
350
400
Magnetics and Energy Conversion
2-240
Today’s Summary
• Today we’ve covered:
• Maxwell’s equations: Ampere’s, Faraday’s and Gauss’
laws
• Soft magnetic materials (steels, etc.)
• Hard magnetic materials (permanent magnets)
• Basic transformers
• The per-unit system
• We started electromechanical conversion basics
Electromechanics
Magnetics and Energy Conversion
2-241